"곡선"의 두 판 사이의 차이
(피타고라스님이 이 페이지를 개설하였습니다.) |
|||
| 1번째 줄: | 1번째 줄: | ||
| + | Let the curve be given by <math>\overrightarrow{r}(t)=(\cos t,\sin t, 3t)</math>. Find the arclength from <math>(1,0,0)</math> to <math>(1,0,6\pi)</math> and the curvature which is constant. | ||
| + | |||
| + | |||
| + | At <math>(1,0,0)</math>, <math>t=0</math> and at <math>(1,0,6\pi)</math>, <math>t=2\pi</math> | ||
| + | |||
| + | <math>\overrightarrow{r}'(t)=(-\sin t,\cos t, 3)</math> | ||
| + | |||
| + | <math>|\overrightarrow{r}'(t)| =\sqrt{\sin^2 t+\cos^2 t +9}=\sqrt{10}</math> | ||
| + | |||
| + | The arclength is given by | ||
| + | |||
| + | <math>L=\int_{0}^{2\pi}|\overrightarrow{r}'(t)| \,dt=\int_{0}^{2\pi}\sqrt{10}\,dt=2\sqrt{10}\pi</math> | ||
| + | |||
| + | <math>\overrightarrow{T}(t)=\frac{\overrightarrow{r}'(t)}{|\overrightarrow{r}'(t)|}=\frac{(-\sin t,\cos t, 3)}{\sqrt{10}}</math> | ||
| + | |||
| + | <math>\overrightarrow{T}'(t)=\frac{(-\cos t,-\sin t, 0)}{\sqrt{10}}</math> | ||
| + | |||
| + | <math>k=\frac{|\overrightarrow{T}'(t)|}{|\overrightarrow{r}'(t)|}=\frac{\frac{|(-\cos t,\sin t, 0)|}{\sqrt{10}}}{\sqrt{10}}=\frac{1}{10}</math> | ||
2009년 12월 31일 (목) 07:37 판
Let the curve be given by \(\overrightarrow{r}(t)=(\cos t,\sin t, 3t)\). Find the arclength from \((1,0,0)\) to \((1,0,6\pi)\) and the curvature which is constant.
At \((1,0,0)\), \(t=0\) and at \((1,0,6\pi)\), \(t=2\pi\)
\(\overrightarrow{r}'(t)=(-\sin t,\cos t, 3)\)
\(|\overrightarrow{r}'(t)| =\sqrt{\sin^2 t+\cos^2 t +9}=\sqrt{10}\)
The arclength is given by
\(L=\int_{0}^{2\pi}|\overrightarrow{r}'(t)| \,dt=\int_{0}^{2\pi}\sqrt{10}\,dt=2\sqrt{10}\pi\)
\(\overrightarrow{T}(t)=\frac{\overrightarrow{r}'(t)}{|\overrightarrow{r}'(t)|}=\frac{(-\sin t,\cos t, 3)}{\sqrt{10}}\)
\(\overrightarrow{T}'(t)=\frac{(-\cos t,-\sin t, 0)}{\sqrt{10}}\)
\(k=\frac{|\overrightarrow{T}'(t)|}{|\overrightarrow{r}'(t)|}=\frac{\frac{|(-\cos t,\sin t, 0)|}{\sqrt{10}}}{\sqrt{10}}=\frac{1}{10}\)